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There are 2 equations a1 and a2, the difference between them is the red words part(integral term).

equation a1 and a2

Boundary x(0)=0,x'(0)=0

x'(t-u) means Bring t-u into the derivative of x ,for example, if x(t)=t^3, then x'(t-u)=3*(t-u)^2

However ,I get the same curve by different codes .so I want to know what's wrong with my code especially the integral term(red words).

The following is code of a1

ieqn = 1 - 6.25*10^5*x[t] + 
1.234*10^4*Integrate[x'[t - u]/Sqrt[u], {u, 0, t}] == 1.5924*x''[t];
ic = {x[0] == 0, x'[0] == 0};
sol = DSolve[{ieqn, ic}, x[t], t];
Plot[x[t] /. sol, {t, 0, 0.005}]

The following is code of a2

ieqn = 1 - 6.25*10^5*x[t] == 1.59236*x''[t];
ic = {x[0] == 0, x'[0] == 0};
sol = DSolve[{ieqn, ic}, x[t], t];
Plot[x[t] /. sol, {t, 0, 0.005}]

same curve This equation is from this paper

The following is the correct curve from this paper.

correct curve of a1:red curve arrowhead

correct curve of a2:dashed black arrowhead

As far as I know, those equation can only have numerical solution.

curve in paper

In the following website, they discussed the same question with mine, but I think there is some wrong.

At last ,I wonder what's wrong with my code especially the integral term(red words).

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  • $\begingroup$ The units in my equations is m and s,and those in figure of paper is um and ms,so the Magnitude of my output is correct $\endgroup$ – Hukai Jul 10 '18 at 3:00
  • 2
    $\begingroup$ What version of Mathematica are you using? $\endgroup$ – bbgodfrey Jul 10 '18 at 5:56
  • $\begingroup$ Your equation is essentially the same as the one in the linked post. Have you tried the method based on Laplace transform in my answer? $\endgroup$ – xzczd Jul 10 '18 at 7:10
  • $\begingroup$ If you are using version 11.0, you are a victim of the bug described in 134465. DSolve instead returns unevaluated in later versions instead of just ignoring the integral. $\endgroup$ – bbgodfrey Jul 10 '18 at 13:20

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